Ordering $8$ people around table if $3$ people do not want to sit next to each other How many ways are there for 8 people to sit around a circular table if none of Alice, Bob, and Eve (three of the 8 people) want to sit next to each other? Two seatings are considered the same if one is a rotation of the other.
I am terrible at these kinds of "sit-uating" problems (haha). My idea is that the three people can be situated in $(3-1)!$ ways and that the rest can be situated in $(5-1)!$ ways. Then, the ordering of the five people depends on the partitions of 5 into 3 groups:

*

*(1,1,3)/(1,3,1)/(3,1,1)

*(1,2,2)/(2,1,2)/(2,2,1)

So is the answer $2!*4!*6=288$? I am not sure if this is right or now, so any help or solutions would be greatly appreciated.
 A: Have Alice take a seat. Have the five people other than Bob and Eve stand around the table in any of $5!$ orders. Have Bob, then Eve, insert themselves, one at a time, between any two of those five people, which they can do in $4\cdot3$ different ways. Then have everyone sit. The total number of arrangements is thus
$$5!\cdot4\cdot3=120\cdot12=1440$$
A: Assume Five other people as  $\ \ \displaystyle x_1,x_2,x_3,x_4,x_5 \ \ $ and the three others who do not wish to sit together as $\ \ \displaystyle a, \ b, \ c. \ \ $
Then for linear arrangement count the cases :
$$\ \ \displaystyle x_1 - x_2 - x_3 - x_4 - x_5 - \ \ $$ Here dashes show spaces in which we can choose any of the three positions for $\ \ \displaystyle a, \ b, \ c. \ \ $
(EDIT: as suggested by N.F. Taussig - This step would determine the number of ways in which neither of $\ \ \displaystyle a, \ b, \ c. \ \ $ are adjacent.)
WAYS: $\ \ \displaystyle \binom {5}{3} \cdot 3! \ \ $
Now arrange rest of the people - WAYS: $\ \ \displaystyle 5! \ \ $
By Multiplication Principle, Total ways of Linear arrangements is
$$\ \ \displaystyle 5! \binom {5}{3} \cdot 3!  \ \ $$
But due to rotational Parity we divide it by $5$. Hence Total circular arrangements:
$$\ \ \displaystyle 4! \binom {5}{3} \cdot 3!  = 1440 \ \ $$
A: I would think about it like this. To eliminate having to worry about over-counting due to rotational symmetry, let's fix Alice's spot on the table.
We run casework on Bob and Eve. WLOG, let's seat Bob first. Notice that there are $8 - 3 = 5$ options to pick from, since Bob cannot occupy the seats next to Alice (and, obviously, Bob cannot take Alice's seat).
Suppose we let Bob sit $2$ spaces counter-clockwise from Alice. Then Eve has $3$ options for where to sit, and for each of those, the remaining $5$ people have $5!$ ways to sit. The same goes for if Bob sits $2$ seats clockwise from Alice (due to reflectional symmetry). This already gives us $2 \times 3 \times 5! = 720$ possibilities.
Now, suppose Bob sits $3$ spaces counter-clockwise from Alice. Eve now only has two options, and each of those options are associated with $5!$ orderings for the other folks. Again, we multiply this count by $2$ because Bob can also sit $3$ spaces clockwise from Alice. This case gives $2 \times 2 \times 5! = 480$ seatings.
Lastly, consider the case in which Bob sits directly opposite of Alice ($4$ spaces in either direction). Again, Eve has $2$ available selections, each of which has $5!$ cases for the other people. This case gives $2 \times 5! = 240$ cases. Note that this case is not doubled because it only accounts for one seating for Bob.
Tallying up, we get $720 + 480 + 240 = \boxed{1440}$ seatings.
(As a sanity check, the total number of ways to seat $8$ people around a circular table, disregarding rotations, is $7! = 5040.$)
